3.1.1 \(\int (b \coth (c+d x))^{7/2} \, dx\) [1]

Optimal. Leaf size=97 \[ \frac {b^{7/2} \text {ArcTan}\left (\frac {\sqrt {b \coth (c+d x)}}{\sqrt {b}}\right )}{d}+\frac {b^{7/2} \tanh ^{-1}\left (\frac {\sqrt {b \coth (c+d x)}}{\sqrt {b}}\right )}{d}-\frac {2 b^3 \sqrt {b \coth (c+d x)}}{d}-\frac {2 b (b \coth (c+d x))^{5/2}}{5 d} \]

[Out]

b^(7/2)*arctan((b*coth(d*x+c))^(1/2)/b^(1/2))/d+b^(7/2)*arctanh((b*coth(d*x+c))^(1/2)/b^(1/2))/d-2/5*b*(b*coth
(d*x+c))^(5/2)/d-2*b^3*(b*coth(d*x+c))^(1/2)/d

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Rubi [A]
time = 0.05, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3554, 3557, 335, 218, 212, 209} \begin {gather*} \frac {b^{7/2} \text {ArcTan}\left (\frac {\sqrt {b \coth (c+d x)}}{\sqrt {b}}\right )}{d}+\frac {b^{7/2} \tanh ^{-1}\left (\frac {\sqrt {b \coth (c+d x)}}{\sqrt {b}}\right )}{d}-\frac {2 b^3 \sqrt {b \coth (c+d x)}}{d}-\frac {2 b (b \coth (c+d x))^{5/2}}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*Coth[c + d*x])^(7/2),x]

[Out]

(b^(7/2)*ArcTan[Sqrt[b*Coth[c + d*x]]/Sqrt[b]])/d + (b^(7/2)*ArcTanh[Sqrt[b*Coth[c + d*x]]/Sqrt[b]])/d - (2*b^
3*Sqrt[b*Coth[c + d*x]])/d - (2*b*(b*Coth[c + d*x])^(5/2))/(5*d)

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[a/b, 0]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 3554

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3557

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rubi steps

\begin {align*} \int (b \coth (c+d x))^{7/2} \, dx &=-\frac {2 b (b \coth (c+d x))^{5/2}}{5 d}+b^2 \int (b \coth (c+d x))^{3/2} \, dx\\ &=-\frac {2 b^3 \sqrt {b \coth (c+d x)}}{d}-\frac {2 b (b \coth (c+d x))^{5/2}}{5 d}+b^4 \int \frac {1}{\sqrt {b \coth (c+d x)}} \, dx\\ &=-\frac {2 b^3 \sqrt {b \coth (c+d x)}}{d}-\frac {2 b (b \coth (c+d x))^{5/2}}{5 d}-\frac {b^5 \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (-b^2+x^2\right )} \, dx,x,b \coth (c+d x)\right )}{d}\\ &=-\frac {2 b^3 \sqrt {b \coth (c+d x)}}{d}-\frac {2 b (b \coth (c+d x))^{5/2}}{5 d}-\frac {\left (2 b^5\right ) \text {Subst}\left (\int \frac {1}{-b^2+x^4} \, dx,x,\sqrt {b \coth (c+d x)}\right )}{d}\\ &=-\frac {2 b^3 \sqrt {b \coth (c+d x)}}{d}-\frac {2 b (b \coth (c+d x))^{5/2}}{5 d}+\frac {b^4 \text {Subst}\left (\int \frac {1}{b-x^2} \, dx,x,\sqrt {b \coth (c+d x)}\right )}{d}+\frac {b^4 \text {Subst}\left (\int \frac {1}{b+x^2} \, dx,x,\sqrt {b \coth (c+d x)}\right )}{d}\\ &=\frac {b^{7/2} \tan ^{-1}\left (\frac {\sqrt {b \coth (c+d x)}}{\sqrt {b}}\right )}{d}+\frac {b^{7/2} \tanh ^{-1}\left (\frac {\sqrt {b \coth (c+d x)}}{\sqrt {b}}\right )}{d}-\frac {2 b^3 \sqrt {b \coth (c+d x)}}{d}-\frac {2 b (b \coth (c+d x))^{5/2}}{5 d}\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 83, normalized size = 0.86 \begin {gather*} \frac {b^3 \sqrt {b \coth (c+d x)} \left (5 \text {ArcTan}\left (\sqrt {\coth (c+d x)}\right )+5 \tanh ^{-1}\left (\sqrt {\coth (c+d x)}\right )-10 \sqrt {\coth (c+d x)}-2 \coth ^{\frac {5}{2}}(c+d x)\right )}{5 d \sqrt {\coth (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*Coth[c + d*x])^(7/2),x]

[Out]

(b^3*Sqrt[b*Coth[c + d*x]]*(5*ArcTan[Sqrt[Coth[c + d*x]]] + 5*ArcTanh[Sqrt[Coth[c + d*x]]] - 10*Sqrt[Coth[c +
d*x]] - 2*Coth[c + d*x]^(5/2)))/(5*d*Sqrt[Coth[c + d*x]])

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Maple [A]
time = 2.19, size = 74, normalized size = 0.76

method	result	size
derivativedivides	\(-\frac {2 b \left (\frac {\left (b \coth \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+b^{2} \sqrt {b \coth \left (d x +c \right )}-\frac {b^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {b \coth \left (d x +c \right )}}{\sqrt {b}}\right )}{2}-\frac {b^{\frac {5}{2}} \arctan \left (\frac {\sqrt {b \coth \left (d x +c \right )}}{\sqrt {b}}\right )}{2}\right )}{d}\)	\(74\)
default	\(-\frac {2 b \left (\frac {\left (b \coth \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+b^{2} \sqrt {b \coth \left (d x +c \right )}-\frac {b^{\frac {5}{2}} \arctanh \left (\frac {\sqrt {b \coth \left (d x +c \right )}}{\sqrt {b}}\right )}{2}-\frac {b^{\frac {5}{2}} \arctan \left (\frac {\sqrt {b \coth \left (d x +c \right )}}{\sqrt {b}}\right )}{2}\right )}{d}\)	\(74\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*coth(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

-2/d*b*(1/5*(b*coth(d*x+c))^(5/2)+b^2*(b*coth(d*x+c))^(1/2)-1/2*b^(5/2)*arctanh((b*coth(d*x+c))^(1/2)/b^(1/2))
-1/2*b^(5/2)*arctan((b*coth(d*x+c))^(1/2)/b^(1/2)))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*coth(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

integrate((b*coth(d*x + c))^(7/2), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 762 vs. \(2 (79) = 158\).
time = 0.37, size = 1574, normalized size = 16.23 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*coth(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

[-1/20*(10*(b^3*cosh(d*x + c)^4 + 4*b^3*cosh(d*x + c)*sinh(d*x + c)^3 + b^3*sinh(d*x + c)^4 - 2*b^3*cosh(d*x +
c)^2 + b^3 + 2*(3*b^3*cosh(d*x + c)^2 - b^3)*sinh(d*x + c)^2 + 4*(b^3*cosh(d*x + c)^3 - b^3*cosh(d*x + c))*si
nh(d*x + c))*sqrt(-b)*arctan((cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2)*sqrt(-b)*sqrt
(b*cosh(d*x + c)/sinh(d*x + c))/(b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2 + b))
- 5*(b^3*cosh(d*x + c)^4 + 4*b^3*cosh(d*x + c)*sinh(d*x + c)^3 + b^3*sinh(d*x + c)^4 - 2*b^3*cosh(d*x + c)^2
+ b^3 + 2*(3*b^3*cosh(d*x + c)^2 - b^3)*sinh(d*x + c)^2 + 4*(b^3*cosh(d*x + c)^3 - b^3*cosh(d*x + c))*sinh(d*x
+ c))*sqrt(-b)*log(-(b*cosh(d*x + c)^4 + 4*b*cosh(d*x + c)^3*sinh(d*x + c) + 6*b*cosh(d*x + c)^2*sinh(d*x + c
)^2 + 4*b*cosh(d*x + c)*sinh(d*x + c)^3 + b*sinh(d*x + c)^4 - 2*(cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x +
c) + sinh(d*x + c)^2 - 1)*sqrt(-b)*sqrt(b*cosh(d*x + c)/sinh(d*x + c)) - 2*b)/(cosh(d*x + c)^4 + 4*cosh(d*x +
c)^3*sinh(d*x + c) + 6*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*cosh(d*x + c)*sinh(d*x + c)^3 + sinh(d*x + c)^4)) +
16*(3*b^3*cosh(d*x + c)^4 + 12*b^3*cosh(d*x + c)*sinh(d*x + c)^3 + 3*b^3*sinh(d*x + c)^4 - 4*b^3*cosh(d*x + c
)^2 + 3*b^3 + 2*(9*b^3*cosh(d*x + c)^2 - 2*b^3)*sinh(d*x + c)^2 + 4*(3*b^3*cosh(d*x + c)^3 - 2*b^3*cosh(d*x +
c))*sinh(d*x + c))*sqrt(b*cosh(d*x + c)/sinh(d*x + c)))/(d*cosh(d*x + c)^4 + 4*d*cosh(d*x + c)*sinh(d*x + c)^3
+ d*sinh(d*x + c)^4 - 2*d*cosh(d*x + c)^2 + 2*(3*d*cosh(d*x + c)^2 - d)*sinh(d*x + c)^2 + 4*(d*cosh(d*x + c)^
3 - d*cosh(d*x + c))*sinh(d*x + c) + d), 1/20*(10*(b^3*cosh(d*x + c)^4 + 4*b^3*cosh(d*x + c)*sinh(d*x + c)^3 +
b^3*sinh(d*x + c)^4 - 2*b^3*cosh(d*x + c)^2 + b^3 + 2*(3*b^3*cosh(d*x + c)^2 - b^3)*sinh(d*x + c)^2 + 4*(b^3*
cosh(d*x + c)^3 - b^3*cosh(d*x + c))*sinh(d*x + c))*sqrt(b)*arctan(sqrt(b)*sqrt(b*cosh(d*x + c)/sinh(d*x + c))
/(b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2 + b)) + 5*(b^3*cosh(d*x + c)^4 + 4*b
^3*cosh(d*x + c)*sinh(d*x + c)^3 + b^3*sinh(d*x + c)^4 - 2*b^3*cosh(d*x + c)^2 + b^3 + 2*(3*b^3*cosh(d*x + c)^
2 - b^3)*sinh(d*x + c)^2 + 4*(b^3*cosh(d*x + c)^3 - b^3*cosh(d*x + c))*sinh(d*x + c))*sqrt(b)*log(2*b*cosh(d*x
+ c)^4 + 8*b*cosh(d*x + c)^3*sinh(d*x + c) + 12*b*cosh(d*x + c)^2*sinh(d*x + c)^2 + 8*b*cosh(d*x + c)*sinh(d*
x + c)^3 + 2*b*sinh(d*x + c)^4 + 2*(cosh(d*x + c)^4 + 4*cosh(d*x + c)*sinh(d*x + c)^3 + sinh(d*x + c)^4 + (6*c
osh(d*x + c)^2 - 1)*sinh(d*x + c)^2 - cosh(d*x + c)^2 + 2*(2*cosh(d*x + c)^3 - cosh(d*x + c))*sinh(d*x + c))*s
qrt(b)*sqrt(b*cosh(d*x + c)/sinh(d*x + c)) - b) - 16*(3*b^3*cosh(d*x + c)^4 + 12*b^3*cosh(d*x + c)*sinh(d*x +
c)^3 + 3*b^3*sinh(d*x + c)^4 - 4*b^3*cosh(d*x + c)^2 + 3*b^3 + 2*(9*b^3*cosh(d*x + c)^2 - 2*b^3)*sinh(d*x + c)
^2 + 4*(3*b^3*cosh(d*x + c)^3 - 2*b^3*cosh(d*x + c))*sinh(d*x + c))*sqrt(b*cosh(d*x + c)/sinh(d*x + c)))/(d*co
sh(d*x + c)^4 + 4*d*cosh(d*x + c)*sinh(d*x + c)^3 + d*sinh(d*x + c)^4 - 2*d*cosh(d*x + c)^2 + 2*(3*d*cosh(d*x
+ c)^2 - d)*sinh(d*x + c)^2 + 4*(d*cosh(d*x + c)^3 - d*cosh(d*x + c))*sinh(d*x + c) + d)]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*coth(d*x+c))**(7/2),x)

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 379 vs. \(2 (79) = 158\).
time = 0.53, size = 379, normalized size = 3.91 \begin {gather*} -\frac {10 \, b^{\frac {7}{2}} \arctan \left (-\frac {\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b}}{\sqrt {b}}\right ) \mathrm {sgn}\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right ) + 5 \, b^{\frac {7}{2}} \log \left ({\left | -\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} + \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b} \right |}\right ) \mathrm {sgn}\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right ) - \frac {16 \, {\left (5 \, {\left (\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b}\right )}^{4} b^{4} \mathrm {sgn}\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right ) - 10 \, {\left (\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b}\right )}^{3} b^{\frac {9}{2}} \mathrm {sgn}\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right ) + 20 \, {\left (\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b}\right )}^{2} b^{5} \mathrm {sgn}\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right ) - 10 \, {\left (\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b}\right )} b^{\frac {11}{2}} \mathrm {sgn}\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right ) + 3 \, b^{6} \mathrm {sgn}\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )\right )}}{{\left (\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b} - \sqrt {b}\right )}^{5}}}{10 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*coth(d*x+c))^(7/2),x, algorithm="giac")

[Out]

-1/10*(10*b^(7/2)*arctan(-(sqrt(b)*e^(2*d*x + 2*c) - sqrt(b*e^(4*d*x + 4*c) - b))/sqrt(b))*sgn(e^(2*d*x + 2*c)
- 1) + 5*b^(7/2)*log(abs(-sqrt(b)*e^(2*d*x + 2*c) + sqrt(b*e^(4*d*x + 4*c) - b)))*sgn(e^(2*d*x + 2*c) - 1) -
16*(5*(sqrt(b)*e^(2*d*x + 2*c) - sqrt(b*e^(4*d*x + 4*c) - b))^4*b^4*sgn(e^(2*d*x + 2*c) - 1) - 10*(sqrt(b)*e^(
2*d*x + 2*c) - sqrt(b*e^(4*d*x + 4*c) - b))^3*b^(9/2)*sgn(e^(2*d*x + 2*c) - 1) + 20*(sqrt(b)*e^(2*d*x + 2*c) -
sqrt(b*e^(4*d*x + 4*c) - b))^2*b^5*sgn(e^(2*d*x + 2*c) - 1) - 10*(sqrt(b)*e^(2*d*x + 2*c) - sqrt(b*e^(4*d*x +
4*c) - b))*b^(11/2)*sgn(e^(2*d*x + 2*c) - 1) + 3*b^6*sgn(e^(2*d*x + 2*c) - 1))/(sqrt(b)*e^(2*d*x + 2*c) - sqr
t(b*e^(4*d*x + 4*c) - b) - sqrt(b))^5)/d

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Mupad [B]
time = 1.58, size = 83, normalized size = 0.86 \begin {gather*} \frac {b^{7/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,\mathrm {coth}\left (c+d\,x\right )}}{\sqrt {b}}\right )}{d}-\frac {2\,b^3\,\sqrt {b\,\mathrm {coth}\left (c+d\,x\right )}}{d}-\frac {2\,b\,{\left (b\,\mathrm {coth}\left (c+d\,x\right )\right )}^{5/2}}{5\,d}-\frac {b^{7/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,\mathrm {coth}\left (c+d\,x\right )}\,1{}\mathrm {i}}{\sqrt {b}}\right )\,1{}\mathrm {i}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*coth(c + d*x))^(7/2),x)

[Out]

(b^(7/2)*atan((b*coth(c + d*x))^(1/2)/b^(1/2)))/d - (2*b^3*(b*coth(c + d*x))^(1/2))/d - (2*b*(b*coth(c + d*x))
^(5/2))/(5*d) - (b^(7/2)*atan(((b*coth(c + d*x))^(1/2)*1i)/b^(1/2))*1i)/d

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